SSC CGL 2024 Tier-I · 19 September 2024, Shift 2 · Quantitative Aptitude · Question 19 of 99

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

From the given numbers A, B, C and D, which number is NOT divisible by 11? A = 712712 B = 177210 C = 64614 D = 756148

(a)D
(b)C
(c)B
(d)A

Answer

Why

Correct — A. Test each number by its alternating digit sum taken from the right: add one digit, subtract the next, and see whether the total is 0 or a multiple of 11.

A = 712712 gives 2 − 1 + 7 − 2 + 1 − 7 = 0, so A is divisible.
B = 177210 gives 0 − 1 + 2 − 7 + 7 − 1 = 0, so B is divisible.
C = 64614 gives 4 − 1 + 6 − 4 + 6 = 11, so C is divisible.

D = 756148 gives 8 − 4 + 1 − 6 + 5 − 7 = −3, which is not a multiple of 11.

D is the number that fails the test, and D is what option (a) names.

Why the others are wrong

(b)
C — C = 64614 passes. Its alternating sum is 4 − 1 + 6 − 4 + 6 = 11, and 11 is itself a multiple of 11, so a non-zero total here still means divisible.
(c)
B — B = 177210 passes on an alternating sum of 0, and the division confirms it: 177210 = 11 × 16110.
(d)
A — A = 712712 passes twice over. Its alternating sum is 0, and any number of the form abcabc equals abc × 1001, where 1001 = 7 × 11 × 13.

Concept

Divisibility by 11 rests on 10 leaving a remainder of −1 on division by 11. The place values therefore alternate in sign — 1, −1, 1, −1 upward from the units — and a number is a multiple of 11 exactly when its alternating digit sum is 0 or a multiple of 11.

That sign pattern is the whole rule. Adding the digits straight tests 3 and 9, not 11, and it is the substitution that costs candidates this mark.

Read the option list carefully: the options are the labels A, B, C and D, not the numbers themselves, so the number D sits behind option (a).

Key facts

A number is divisible by 11 when its alternating digit sum is 0 or a multiple of 11.
The rule works because 10 is congruent to −1 modulo 11.
1001 = 7 × 11 × 13, so every six-digit number of the form abcabc is divisible by 7, 11 and 13.
64614 has alternating sum 11, which is divisible, so a non-zero total does not mean failure.

Study next

Divisibility by 7, 11 and 13 together through the factor 1001Finding the missing digit that makes a number divisible by 11Remainder when a large number is divided by 11, read off the alternating sum

Common traps

Adding the digits, which is the test for 3 and 9, instead of alternating the signs.
Calling 64614 indivisible because its alternating sum is 11 rather than 0.
Picking the option lettered d because the failing number is labelled D.

SSC works the 11-test from both ends. It prints four numbers and asks which one fails on 12 Sep 2024, 12:30, Quant Q.21.

Elsewhere it hides a digit and asks you to supply it — 6954P divisible by 11 on 17 Sep 2024, 16:00, Quant Q.10. The same alternating sum answers either version.

Related PYQs

No directly related past PYQ was found.